Question

The charges -2q, 4q, -5q, and 2q are placed at the following locations on the x-axis: —8.00(negative 8.00 m), -3.00 (negative 3.00 m), 3.00, and 8.00 m along the x-axis

(set the origin at 0.00). X-axis are given in meters. Find the total electric field at x = -10.00 m, given that q = 5.00 nC

.

Answer 1

Electric field due to point charge is given by

E=KQ/r²

Total Electric field at X=-10 m is

$E=\frac{K|2q|}{(-10-(-8))^{2}}-\frac{K|4q|}{(-10-(-3))^{2}}+\frac{K|5q|}{(-10-3)^{2}}-\frac{K|2q|}{(-10-8)^{2}}$

E=0.44178Kq =0.44178*(9*10⁹)(5*10^-9)

E=19.88 N/C